Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 22}{x + 8} = \dfrac{-2x + 26}{x + 8}$
Multiply both sides by $x + 8$ $ \dfrac{x^2 - 22}{x + 8} (x + 8) = \dfrac{-2x + 26}{x + 8} (x + 8)$ $ x^2 - 22 = -2x + 26$ Subtract $-2x + 26$ from both sides: $ x^2 - 22 - (-2x + 26) = -2x + 26 - (-2x + 26)$ $ x^2 - 22 + 2x - 26 = 0$ $ x^2 - 48 + 2x = 0$ Factor the expression: $ (x + 8)(x - 6) = 0$ Therefore $x = -8$ or $x = 6$ At $x = -8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -8$, it is an extraneous solution.